| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2004 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Repeated trials until stopping condition |
| Difficulty | Moderate -0.3 This is a straightforward tree diagram problem with clearly stated probabilities and standard conditional probability calculation. While it requires careful organization across multiple branches (first serve, second serve, win/lose outcomes) and application of Bayes' theorem in part (iii), all steps follow routine procedures with no conceptual surprises. The multi-part structure and conditional probability element elevate it slightly above pure recall, but it remains easier than a typical A-level question due to its mechanical nature and explicit guidance. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables |
| Answer | Marks | Guidance |
|---|---|---|
| Tree diagram: top branches \((0.65, 0.9, 0.1)\) | B1 | For top branches correct \((0.65, 0.9, 0.1)\) |
| Bottom branches \((0.35, 0.8, 0.2)\) | B1 | For bottom branches correct \((0.35, 0.8, 0.2)\) |
| Win/Lose after \(2^{\text{nd}}\) in \((0.6, 0.4)\) | B1 | For win/lose option after \(2^{\text{nd}}\) in \((0.6, 0.4)\) |
| All labels including final lose at end of bottom branch | B1 | For all labels including final lose at end of bottom branch |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.65\times0.1 + 0.35\times0.8\times0.4 + 0.35\times2 = 0.247\) | M1 | For evaluating \(1^{\text{st}}\) in and lose seen |
| M1 | For \(1^{\text{st}}\) out \(2^{\text{nd}}\) in lose, or \(1^{\text{st}}\) out \(2^{\text{nd}}\) out lose | |
| A1 | For correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{0.65\times0.1}{0.247} = 0.263\ \left(=\frac{5}{19}\right)\) | M1 | For dividing their \(1^{\text{st}}\) in and lose by their answer to (ii) |
| A1ft | For correct answer; ft only on \(0.65\times0.1/\text{their (ii)}\) |
# Question 6:
## Part (i)
Tree diagram: top branches $(0.65, 0.9, 0.1)$ | B1 | For top branches correct $(0.65, 0.9, 0.1)$
Bottom branches $(0.35, 0.8, 0.2)$ | B1 | For bottom branches correct $(0.35, 0.8, 0.2)$
Win/Lose after $2^{\text{nd}}$ in $(0.6, 0.4)$ | B1 | For win/lose option after $2^{\text{nd}}$ in $(0.6, 0.4)$
All labels including final lose at end of bottom branch | B1 | For all labels including final lose at end of bottom branch
**Total: 4**
## Part (ii)
$0.65\times0.1 + 0.35\times0.8\times0.4 + 0.35\times2 = 0.247$ | M1 | For evaluating $1^{\text{st}}$ in and lose seen
| M1 | For $1^{\text{st}}$ out $2^{\text{nd}}$ in lose, or $1^{\text{st}}$ out $2^{\text{nd}}$ out lose
| A1 | For correct answer
**Total: 3**
## Part (iii)
$\frac{0.65\times0.1}{0.247} = 0.263\ \left(=\frac{5}{19}\right)$ | M1 | For dividing their $1^{\text{st}}$ in and lose by their answer to (ii)
| A1ft | For correct answer; ft only on $0.65\times0.1/\text{their (ii)}$
**Total: 2**
---6 When Don plays tennis, $65 \%$ of his first serves go into the correct area of the court. If the first serve goes into the correct area, his chance of winning the point is $90 \%$. If his first serve does not go into the correct area, Don is allowed a second serve, and of these, $80 \%$ go into the correct area. If the second serve goes into the correct area, his chance of winning the point is $60 \%$. If neither serve goes into the correct area, Don loses the point.\\
(i) Draw a tree diagram to represent this information.\\
(ii) Using your tree diagram, find the probability that Don loses the point.\\
(iii) Find the conditional probability that Don's first serve went into the correct area, given that he loses the point.
\hfill \mbox{\textit{CAIE S1 2004 Q6 [9]}}